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Expand the Einstein tensor about the Minkowski metric, keeping a second-order metric perturbation as well:Here $G^{(2)}[h,h]$ includes all quadratic terms from the Einstein tensor, including the inverse-metric corrections in its contractions. In vacuum, the first-order Linearized Einstein equations give $G^{(1)}[h]=0$, while the next order gives $G^{(1)}[j]=-G^{(2)}[h,h]$. Define the quadratic gravitational effective stress-energy tensor byIt is symmetric because the Einstein tensor is symmetric, and is quadratic in $h$ and its derivatives. The linearized contracted Bianchi identity is the off-shell identity ${\partial }^{\mu }{G}_{\mu \nu }^{(1)}[j]=0$. The second-order vacuum equation therefore impliesOne can obtain the same conclusion directly by expanding the full contracted Bianchi identity to second order: every correction to the flat divergence multiplies $G^{(1)}[h]$ and vanishes on the first-order vacuum solution. Thus conservation holds for vacuum solutions of linearized gravity, rather than for an arbitrary off-shell $h$. Relative to the fixed Minkowski background, $t$ is a tensor under background changes of coordinates, but this does not make it invariant under the perturbative gauge freedom.

The Einstein tensor is $G_{\alpha \beta }=R_{\alpha \beta }-\frac{1}{2}{g}_{\alpha \beta }R$. Taking the metric trace of the Einstein field equations in four dimensions givesand substitution back gives the Trace-reversed Einstein field equations in D dimensions specialized to $D=4$:For linearized gravity, the given equation and $G_{\mu \nu }=8\pi T_{\mu \nu }$ implySince $\bar{h}=-h$, its trace is $\square h=16\pi T$. Using ${\bar{h}}_{\mu \nu }=h_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }h$ then yields